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%\author{王立庆（2019级数学与应用数学1班）}
\author{学号 \underline{\hspace{4cm}} 姓名  \underline{\hspace{4cm}} }
%\title{高等代数第六章：向量空间}
\title{第九章二次型考试解答 }
%\date{\vspace{-3ex}}
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\date{2023年5月31日}

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\begin{document}

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%\begin{abstract}
%%主要内容：
%7.3. 
%7.4. 
%7.5. 

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\begin{enumerate}

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\item %第1题
设实二次型 $q(x_1, x_2, x_3)=x_1^2+2x_2^2+x_3^2+2x_1x_2+6x_1x_3+8x_2x_3$. 
\begin{enumerate}
\item  使用配方法，将 $q$ 化为实二次型的标准型，写出使用的变量代换。
\item  求实二次型 $q$ 的秩与符号差，判断 $q$ 是否为正定二次型。
\item  求实对称阵 $A$ 使得 $q(x_1,x_2,x_3)=X^tAX$. 
\item  计算矩阵 $A$ 的顺序主子式，判断 $A$ 是否为正定矩阵。
\end{enumerate}

\vspace{0.2cm}

{\color{red} 解答：
\begin{enumerate}

\item  先将与 $x_1$ 有关的项配成完全平方，然后在剩下的项中，将与 $x_2$ 有关的项配成完全平方，可得
\begin{eqnarray*}
q(x_1, x_2, x_3) &=& x_1^2+2x_2^2+x_3^2+2x_1x_2+6x_1x_3+8x_2x_3 \\ 
&=& (x_1+x_2+3x_3)^2 + x_2^2+2x_2x_3 - 8x_3^2 \\
&=& (x_1+x_2+3x_3)^2 + (x_2+x_3)^2 - 9x_3^2 \\ 
&=& y_1^2+y_2^2-y_3^2. 
\end{eqnarray*}
使用的变量代换为
\begin{eqnarray*}
\left\{\begin{array}{rcl}
y_1 &=&  x_1+x_2+3x_3 \\ 
y_2 &=& x_2+x_3 \\ 
y_3 &=& 3x_3.
\end{array}\right.
\end{eqnarray*}

\item  因为实二次型 $y_1^2+y_2^2-y_3^2$ 的正惯性指数 $p=2$, 负惯性指数 $q=1$, 所以秩 $r=p+q=3$, 符号差 $s=p-q=1$. 这个实二次型不是正定的。

\item  将这个二次型写成矩阵乘积的形式，可得
\begin{eqnarray*}
x_1^2+2x_2^2+x_3^2+2x_1x_2+6x_1x_3+8x_2x_3 = 
\begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}
\begin{bmatrix} 1 & 1 &3  \\ 1 & 2 &4 \\  3& 4 &1  \end{bmatrix}
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}. 
\end{eqnarray*}
于是所求的对称矩阵为 
\begin{eqnarray*}
A=\begin{bmatrix} 1 & 1 &3  \\ 1 & 2 &4 \\  3& 4 &1  \end{bmatrix}. 
\end{eqnarray*}

\item  矩阵 $A$ 的顺序主子式为
\begin{eqnarray*}
A_1 = 1, \hspace{0.3cm} 
A_2 = \begin{vmatrix} 1 & 1  \\ 1 & 2  \end{vmatrix}=1,  \hspace{0.3cm} 
A_3 = \begin{vmatrix} 1 & 1 &3  \\ 1 & 2 &4 \\  3& 4 &1  \end{vmatrix} = -9. 
\end{eqnarray*}
因为矩阵 $A$ 的顺序主子式不是都大于零，所以 $A$ 不是正定矩阵。

\end{enumerate}

}

\vspace{0.2cm}


\newpage

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\item  %第2题
设实二次型 $q(x_1,x_2)=-2x_1^2+8x_1x_2-2x_2^2$. 
\begin{enumerate}
\item  求实对称阵 $A$ 使得 $q(x_1,x_2)=X^tAX$. 
\item  求正交矩阵 $U$ 使得 $U^tAU$ 为对角矩阵。
\item  画出二次曲线 $q(x_1,x_2)=4$ 的图像，写出主轴的方程。
\end{enumerate}

\vspace{0.2cm}

{\color{red} 解答：
\begin{enumerate}
\item  将这个二次型写成矩阵乘积的形式，可得
\begin{eqnarray*}
q(X) = -2x_1^2 +8x_1x_2 -2x_2^2 = 
\begin{bmatrix} x_1 & x_2 \end{bmatrix}
\begin{bmatrix} -2 & 4 \\ 4 & -2 \end{bmatrix}
\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
=X^tAX. 
\end{eqnarray*}

\item  按照实对称阵正交相似于对角矩阵的步骤进行计算。
\begin{enumerate}
\item  计算矩阵 $A$ 的特征值。根据 $|\lambda E - A| = \lambda^2 + 4\lambda -12$, 得到两个特征值 $\lambda_1=2, \lambda_2=-6$.  

\item  根据齐次线性方程组 $(\lambda E-A)\alpha=0$ 计算特征向量。
属于特征值 $\lambda_1=2$ 的特征向量为 $\alpha_1=k(1,1)^t$. 
属于特征值 $\lambda_2=-6$ 的特征向量为 $\alpha_2=k(1,-1)^t$.  

\item  因为对称矩阵的属于不同的特征值的特征向量已经相互正交，所以直接将它们单位化，可得 $\eta_1=\frac{1}{\sqrt{2}}(1,1)^t$ 与 $\eta_2=\frac{1}{\sqrt{2}}(1,-1)^t$. 

\item  将这些单位化的相互正交的特征向量按照列向量的方式排列成一个矩阵，可得所求的正交矩阵为
\begin{eqnarray*}
U=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}.  
\end{eqnarray*} 
这个矩阵不是唯一确定的，两个列向量可以交换次序，每个列向量可以相差一个负号。

\item  
最后矩阵 $A$ 正交相似于由特征值作为对角线元素的对角阵，即有 
\begin{eqnarray*}
U^{-1}AU = U^tAU = \begin{bmatrix} 2 & 0 \\ 0 & -6 \end{bmatrix}. 
\end{eqnarray*}

\end{enumerate}

\item  在老坐标系中画出新坐标系。在新坐标系中画出图像，写出主轴方程。在老坐标系中写出主轴方程。

\begin{enumerate}

\item  取变量的正交变换 $X=UY$. 所以 $Y=U^{-1}X=U^tX$, 即
\begin{eqnarray*}
\begin{bmatrix} y_1 \\ y_2 \end{bmatrix}
= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}
\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. 
\end{eqnarray*}
这时，新老坐标系的关系为
\begin{eqnarray*}
\left\{\begin{array}{rcl}
y_1 &=&  \frac{1}{\sqrt{2}}x_1+ \frac{1}{\sqrt{2}}x_2, \\ 
y_2 &=& \frac{1}{\sqrt{2}} x_1- \frac{1}{\sqrt{2}}x_2.
\end{array}\right.
\end{eqnarray*}

\item  题目所给的实二次型在变量代换后化为
\begin{eqnarray*}
q(X)=q(UY) = (UY)^tA(UY) = Y^t U^t AUY 
= \begin{bmatrix} y_1 & y_2 \end{bmatrix} 
\begin{bmatrix} 2 & 0 \\ 0 & -6 \end{bmatrix}
\begin{bmatrix} y_1 \\ y_2 \end{bmatrix}
=2y_1^2-6y_2^2. 
\end{eqnarray*}

\item  因此 $q(X)=4$ 在新坐标系中为 $2y_1^2-6y_2^2=4$. 这是双曲线。主轴方程为
$y_2=0$ 与 $y_1=0$. 

\item  所以在老坐标系中，主轴方程为 $x_1-x_2=0$ 与 $x_1+x_2=0$. 

\begin{center}
\includegraphics[height=5cm, width=5cm]{exercise_parabola.png}
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\end{enumerate}

\end{enumerate}

}

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\end{enumerate}


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